A body is thrown with initial velocity 10 m/sec. at an angle `37^(@)` fro horizontal. Find
(i) time of flight
(ii) Maximum geight.
(iii) Range (iv) Position vector after `t=1` sec.

Explanation

To solve the problem step by step, let's break it down into the four parts as requested:

### Given:
- Initial velocity, \( u = 10 \, \text{m/s} \)
- Angle of projection, \( \theta = 37^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### (i) Time of Flight
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
Substituting the values:
\[ T = \frac{2 \times 10 \times \sin(37^\circ)}{10} \]
\[ T = 2 \times \sin(37^\circ) \]
Using the value \( \sin(37^\circ) = \frac{3}{5} \):
\[ T = 2 \times \frac{3}{5} = \frac{6}{5} = 1.2 \, \text{s} \]

### (ii) Maximum Height
The maximum height \( H \) is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Substituting the values:
\[ H = \frac{10^2 \times \sin^2(37^\circ)}{2 \times 10} \]
\[ H = \frac{100 \times \left(\frac{3}{5}\right)^2}{20} \]
\[ H = \frac{100 \times \frac{9}{25}}{20} = \frac{900}{500} = 1.8 \, \text{m} \]

### (iii) Range
The range \( R \) of the projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Using \( \sin(2\theta) = \sin(74^\circ) \):
\[ R = \frac{10^2 \times \sin(74^\circ)}{10} \]
Using \( \sin(74^\circ) \approx 0.961 \):
\[ R = \frac{100 \times 0.961}{10} = 9.61 \, \text{m} \]

### (iv) Position Vector after \( t = 1 \, \text{s} \)
To find the position vector after \( t = 1 \, \text{s} \), we need to calculate the horizontal and vertical displacements.

**Horizontal Displacement \( x \)**:
\[ x = u \cos \theta \cdot t \]
Calculating \( u \cos(37^\circ) \):
\[ u \cos(37^\circ) = 10 \times \frac{4}{5} = 8 \, \text{m/s} \]
Thus,
\[ x = 8 \cdot 1 = 8 \, \text{m} \]

**Vertical Displacement \( y \)**:
\[ y = u \sin \theta \cdot t - \frac{1}{2} g t^2 \]
Calculating \( u \sin(37^\circ) \):
\[ u \sin(37^\circ) = 10 \times \frac{3}{5} = 6 \, \text{m/s} \]
Thus,
\[ y = 6 \cdot 1 - \frac{1}{2} \cdot 10 \cdot 1^2 = 6 - 5 = 1 \, \text{m} \]

**Position Vector**:
The position vector \( \vec{r} \) after \( t = 1 \, \text{s} \) is given by:
\[ \vec{r} = x \hat{i} + y \hat{j} = 8 \hat{i} + 1 \hat{j} \, \text{m} \]

### Summary of Results:
1. Time of Flight: \( 1.2 \, \text{s} \)
2. Maximum Height: \( 1.8 \, \text{m} \)
3. Range: \( 9.61 \, \text{m} \)
4. Position Vector after \( t = 1 \, \text{s} \): \( \vec{r} = 8 \hat{i} + 1 \hat{j} \, \text{m} \)